Codeforces Round #479 (Div. 3)题解
2024-10-09 03:01:34
CF首次推出div3给我这种辣鸡做，当然得写份博客纪念下
A. Wrong Subtraction
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one;
- if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number nn. Tanya will subtract one from it kk times. Your task is to print the result after all kk subtractions.
It is guaranteed that the result will be positive integer number.
Input
The first line of the input contains two integer numbers nn and kk (2≤n≤1092≤n≤109, 1≤k≤501≤k≤50) — the number from which Tanya will subtract and the number of subtractions correspondingly.
Output
Print one integer number — the result of the decreasing nn by one kk times.
It is guaranteed that the result will be positive integer number.
Examples
input
Copy
512 4
output
Copy
50
input
Copy
1000000000 9
output
Copy
1
Note
The first example corresponds to the following sequence: 512→511→510→51→50512→511→510→51→50.
分析：水题，就是末尾是0就除10，不为0就-1
#include <bits/stdc++.h>
using namespace std; #define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 1001
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; int main(void){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie();
int i, n, k; cin>>n>>k;
for(i = ; i < k; i++){
if(n % != ) n -= ;
else n /= ;
}
cout<<n<<endl;
return EXIT_SUCCESS;
}
B. Two-gram
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string ss consisting of nn capital Latin letters. Your task is to find any two-gram contained in the given string as a substring(i.e. two consecutive characters of the string) maximal number of times. For example, for string ss = "BBAABBBA" the answer is two-gram "BB", which contained in ss three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number nn (2≤n≤1002≤n≤100) — the length of string ss. The second line of the input contains the string ss consisting of nn capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string ss as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
input
Copy
7
ABACABA
output
Copy
AB
input
Copy
5
ZZZAA
output
Copy
ZZ 分析：水题，二维哈希映射可以做，我是直接string.find()查的
#include <bits/stdc++.h>
using namespace std; #define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 1001
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; int main(void){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie();
int i, m, len, loca = - , max = , num[MAX] = {}, flag; string str, str1; cin>>len>>str;
for(i = ; i < len - ; i++){
str1 = "";m = ;str1 += str[i]; str1 += str[i + ];
flag = str.find(str1, );
while(flag != string::npos){num[i]++;flag = str.find(str1, flag + );}
}
for(i = ; i < len - ; i++)if(num[i] > max){loca = i;max = num[i];} cout<<str[loca]<<str[loca + ]<<endl; return EXIT_SUCCESS;
}
C. Less or Equal
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a sequence of integers of length nn and integer number kk. You should print any integer number xx in the range of [1;109][1;109] (i.e. 1≤x≤1091≤x≤109) such that exactly kk elements of given sequence are less than or equal to xx.
Note that the sequence can contain equal elements.
If there is no such xx, print "-1" (without quotes).
Input
The first line of the input contains integer numbers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, 0≤k≤n0≤k≤n). The second line of the input contains nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the sequence itself.
Output
Print any integer number xx from range [1;109][1;109] such that exactly kk elements of given sequence is less or equal to xx.
If there is no such xx, print "-1" (without quotes).
Examples
input
Copy
7 4
3 7 5 1 10 3 20
output
Copy
6
input
Copy
7 2
3 7 5 1 10 3 20
output
Copy
-1
Note
In the first example 55 is also a valid answer because the elements with indices [1,3,4,6][1,3,4,6] is less than or equal to 55 and obviously less than or equal to 66.
In the second example you cannot choose any number that only 22 elements of the given sequence will be less than or equal to this number because 33 elements of the given sequence will be also less than or equal to this number.
分析：找x，大于等于num[]里面的k个元素,3个条件
1.如果k == 0且，num[1]>1（就是没有x大于num里面的0个元素）所以输入1，如果k==0且num[1] <= 1,说明
这时x还可以大于num里面的元素，与题意不符，所以输出-1。
2.k == n，x要大于等于num[]里面所有元素，题目中num[]最大为1000000000。
3.0 < k < n，只需要num[k] != num[k + 1]即可，因为这样就有k + 1个元素了，而题目只要k个元素
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#include <bits/stdc++.h>
using namespace std;
#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 220000
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
int main(void){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie();
int i, n, k, num[MAX], ans, id;
cin>>n>>k;
for(i = ; i <= n; i++) cin>>num[i];
sort(num + , num + n + );
if(k == ){
if(num[] > ) cout<<<<endl;
else cout<<-<<endl;
}
else if(n == k) cout<<<<endl;
else{
if(num[k] == num[k + ]) cout<<-<<endl;
else cout<<num[k]<<endl;
}
return EXIT_SUCCESS;
}
D. Divide by three, multiply by two
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, and then performs with it n−1n−1operations of the two kinds:
- divide the number xx by 33 (xx must be divisible by 33);
- multiply the number xx by 22.
After each operation, Polycarp writes down the result on the board and replaces xx by the result. So there will be nn numbers on the board after all.
You are given a sequence of length nn — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.
Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.
It is guaranteed that the answer exists.
Input
The first line of the input contatins an integer number nn (2≤n≤1002≤n≤100) — the number of the elements in the sequence. The second line of the input contains nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤3⋅10181≤ai≤3⋅1018) — rearranged (reordered) sequence that Polycarp can wrote down on the board.
Output
Print nn integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.
It is guaranteed that the answer exists.
Examples
input
Copy
6
4 8 6 3 12 9
output
Copy
9 3 6 12 4 8
input
Copy
4
42 28 84 126
output
Copy
126 42 84 28
input
Copy
2
1000000000000000000 3000000000000000000
output
Copy
3000000000000000000 1000000000000000000
Note
In the first example the given sequence can be rearranged in the following way: [9,3,6,12,4,8][9,3,6,12,4,8]. It can match possible Polycarp's game which started with x=9x=9.
分析：题目好理解，后一个数是前一个数的 1/ 3，或者2倍,先统计所有数%3的个数，先按照这个数排列，再按照他们%2个数排列（或他们自身数大小排列，一样的，因为是2倍，后面数肯定大于前面数）
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. **/
#include <bits/stdc++.h>
using namespace std;
#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 110
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
struct Node{
LL a;
int b;
}num[MAX];
//ÔªËصÄÖØÅÅÁÐ
bool cmp(const Node& a, const Node& b){
if(a.b != b.b) return a.b > b.b;
else return a.a < b.a;
}
int main(void){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie();
int i, n; LL cnt;
cin>>n;
for(i = ; i < n; i++){
cin>>num[i].a;
cnt = num[i].a;
while(cnt % == ){cnt /= ; num[i].b++;}
}
sort(num, num + n, cmp);
for(i = ; i < n; i++) printf("%lld ", num[i].a);
puts("");
return EXIT_SUCCESS;
}
E. Cyclic Components
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given an undirected graph consisting of nn vertices and mm edges. Your task is to find the number of connected components which are cycles.
Here are some definitions of graph theory.
An undirected graph consists of two sets: set of nodes (called vertices) and set of edges. Each edge connects a pair of vertices. All edges are bidirectional (i.e. if a vertex aa is connected with a vertex bb, a vertex bb is also connected with a vertex aa). An edge can't connect vertex with itself, there is at most one edge between a pair of vertices.
Two vertices uu and vv belong to the same connected component if and only if there is at least one path along edges connecting uu and vv.
A connected component is a cycle if and only if its vertices can be reordered in such a way that:
- the first vertex is connected with the second vertex by an edge,
- the second vertex is connected with the third vertex by an edge,
- ...
- the last vertex is connected with the first vertex by an edge,
- all the described edges of a cycle are distinct.
A cycle doesn't contain any other edges except described above. By definition any cycle contains three or more vertices.
There are 66 connected components, 22 of them are cycles: [7,10,16][7,10,16] and [5,11,9,15][5,11,9,15].
Input
The first line contains two integer numbers nn and mm (1≤n≤2⋅1051≤n≤2⋅105, 0≤m≤2⋅1050≤m≤2⋅105) — number of vertices and edges.
The following mm lines contains edges: edge ii is given as a pair of vertices vivi, uiui (1≤vi,ui≤n1≤vi,ui≤n, ui≠viui≠vi). There is no multiple edges in the given graph, i.e. for each pair (vi,uivi,ui) there no other pairs (vi,uivi,ui) and (ui,viui,vi) in the list of edges.
Output
Print one integer — the number of connected components which are also cycles.
Examples
input
Copy
5 4
1 2
3 4
5 4
3 5
output
Copy
1
input
Copy
17 15
1 8
1 12
5 11
11 9
9 15
15 5
4 13
3 13
4 3
10 16
7 10
16 7
14 3
14 4
17 6
output
Copy
2
Note
In the first example only component [3,4,5][3,4,5] is also a cycle.
The illustration above corresponds to the second example.
分析：题意就是统计环的个数且这个环任意一点只能有2个边,简单DFS图论题目，2维数组统计每个点的度，条件如下：
1.该点度不为2，退出。
2.该点没访问过，标记，并访问与该点关联的所有点，ans++
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OBHs;. :;;,v H: :6X; rpL;7pSrrr7r;::,....,,:;iirrvvvr, :BBPs;;:,,,,::;::
::rPDEL:..7:c;7r pK .: :KL.:r:.. ..:icsS5sr:,. JgGgBQDOSJss77r
:r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs
.:J: :; .,:K6BQS7:.,.,.
:r7Ji;r7;::;;.
. **/
#include <bits/stdc++.h>
using namespace std;
#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 220000
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
vector<int>v[MAX];
int dis[MAX] = {};
int flag;
void DFS(int u){
dis[u] = ; //该点访问
if(v[u].size() != ) flag = false; //若改点的度不为2，则可以直接退出
for(int i = ; i < v[u].size(); i++){ if(!dis[v[u][i]]) DFS(v[u][i]);}
}
int main(void){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie();
int i, j, n, m, a, b, ans;
cin>>n>>m;ans = ;
for(i = ; i <= m; i++){
cin>>a>>b;
//所有点的出入度都记录
v[a].push_back(b);
v[b].push_back(a);
}
for(i = ; i <= n; i++){
if(!dis[i]){
flag = ;DFS(i);
if(flag) ans++;
}
}
cout<<ans<<endl;
return EXIT_SUCCESS;
}
F. Consecutive Subsequence
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given an integer array of length nn.
You have to choose some subsequence of this array of maximum length such that this subsequence forms a increasing sequence of consecutive integers. In other words the required sequence should be equal to [x,x+1,…,x+k−1][x,x+1,…,x+k−1] for some value xx and length kk.
Subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. You can erase any elements, not necessarily going successively. The remaining elements preserve their order. For example, for the array [5,3,1,2,4][5,3,1,2,4] the following arrays are subsequences: [3][3], [5,3,1,2,4][5,3,1,2,4], [5,1,4][5,1,4], but the array [1,3][1,3] is not.
Input
The first line of the input containing integer number nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of the array. The second line of the input containing nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the array itself.
Output
On the first line print kk — the maximum length of the subsequence of the given array that forms an increasing sequence of consecutive integers.
On the second line print the sequence of the indices of the any maximum length subsequence of the given array that forms an increasing sequence of consecutive integers.
Examples
input
Copy
7
3 3 4 7 5 6 8
output
Copy
4
2 3 5 6
input
Copy
6
1 3 5 2 4 6
output
Copy
2
1 4
input
Copy
4
10 9 8 7
output
Copy
1
1
input
Copy
9
6 7 8 3 4 5 9 10 11
output
Copy
6
1 2 3 7 8 9
Note
All valid answers for the first example (as sequences of indices):
- [1,3,5,6][1,3,5,6]
- [2,3,5,6][2,3,5,6]
All valid answers for the second example:
- [1,4][1,4]
- [2,5][2,5]
- [3,6][3,6]
All valid answers for the third example:
- [1][1]
- [2][2]
- [3][3]
- [4][4]
All valid answers for the fourth example:
- [1,2,3,7,8,9]
分析：比LIS更简单，题意就是要找 1， 2， 3， 4， 5这样的单调递增数，不能是1， 2， 3 ， 5， 6, 直接把每个数 = 上一个数 + 1,然后最大数就是个数，然后index = index - ans + 1就是开始的索引,每有一个
index = num[i],index++,输出i即可
/**
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.:J: :; .,:K6BQS7:.,.,.
:r7Ji;r7;::;;.
. **/
#include <bits/stdc++.h>
using namespace std;
#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 220000
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
map<LL, LL> mp;
int main(void){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie();
int i, j, n;LL a[MAX], ans, index;
cin>>n;ans = index = ;
for(i = ; i <= n; i++){cin>>a[i]; mp[a[i]] = mp[a[i] - ] + ;}
for(i = ; i <= n; i++){
if(mp[a[i]] > ans){
ans = mp[a[i]];
index = a[i];
}
}
cout<<ans<<endl;
index = index - ans + ;
for(i = ; i <= n; i++){
if(a[i] == index){
printf("%d ", i);
index++;
}
}
puts("");
return EXIT_SUCCESS;
}
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